Prove $\lim\limits_{(x,y)\rightarrow(0,0)}y\sin(1/x)=0$
We know that $|\sin(1/x)|\leq1$
We need to prove for any $\varepsilon>0$ and $\delta>0$, $|y\sin(1/x)|<\varepsilon$ when $0<\sqrt{x^2+y^2}<\delta$.
We can look at this intuitively, if $y=mx$ , $y\sin(1/x)=mx\sin(1/x)$. The limit can only exist if $x\sin(1/x)=0$ at $0$.
Since $0<\sqrt{x^2+y^2}<\delta$ , $0<y<\sqrt{\delta^2-x^2}$. So $0<y\sin|1/x|<\sqrt{\delta^2-x^2}\sin|1/x|$. I don't know how to proceed further, please help.
Your observation of $|\sin(1/x)|\le 1$ should have almost had everything done. Simply note that $$ 0\le |y|\le\sqrt{x^2+y^2} .$$
So if $\sqrt{x^2+y^2}<\epsilon$, then $$ 0\le |y\sin(1/x)|\le |y|<\epsilon. $$